goroutines是如何工作的,当主要过程结束时它们会死吗？

Question

我编写了一个简单的小例子,将1000万条记录插入mongodb.我开始按顺序工作.然后我查找了如何进行并发,并找到了goroutines.这看起来像我想要的,但它并不像我期望的那样表现.我实现了一个WaitGroup来阻止程序在所有goroutine完成之前退出,但我仍然遇到问题.

所以我将从正在发生的事情开始然后显示代码.当我运行没有goroutine的代码时,所有1000万条记录都插入mongodb罚款.然而,当我添加goroutine时,一些不确定的数量被输入..一般在8500左右给予或采取几百.我检查了mongodb日志,看它是否有问题,没有任何显示.所以我不确定是什么,可能是,只是没有被记录.无论如何,这是代码:

(旁注:我一次只做1条记录,但我把它拆分成一种方法,所以我可以在将来一次测试多条记录......只是还没弄明白如何用mongodb做到这一点然而.)

package main

import (
  "fmt"
  "labix.org/v2/mgo"
  "strconv"
  "time"
  "sync"
)

// structs
type Reading struct {
  Id   string
  Name string
}

var waitGroup sync.WaitGroup

// methods
func main() {
  // Setup timer
  startTime := time.Now()

  // Setup collection
  collection := getCollection("test", "readings")
  fmt.Println("collection complete: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))

  // Setup readings
  readings := prepareReadings()
  fmt.Println("readings prepared: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))

  // Insert readings
  for i := 1; i <= 1000000; i++ {
    waitGroup.Add(1)
    go insertReadings(collection, readings)

    // fmt.Print(".")

    if i % 1000 == 0 {
      fmt.Println("1000 readings queued for insert: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))
    }
  }
  waitGroup.Wait()

  fmt.Println("all readings inserted: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))
}

func getCollection(databaseName string, tableName string) *mgo.Collection {
  session, err := mgo.Dial("localhost")

  if err != nil {
    // panic(err)
    fmt.Println("error getCollection:", err)
  }

  // defer session.Close()

  // Optional. Switch the session to a monotonic behavior.
  // session.SetMode(mgo.Monotonic, true)

  collection := session.DB(databaseName).C(tableName)

  return collection
}

func insertReadings(collection *mgo.Collection, readings []Reading) {
  err := collection.Insert(readings)

  if err != nil {
    // panic(err)
    fmt.Println("error insertReadings:", err)
  }

  waitGroup.Done()
}

func prepareReadings() []Reading {
  var readings []Reading
  for i := 1; i <= 1; i++ {
    readings = append(readings, Reading{Name: "Thing"})
  }

  return readings
}

Answer 1

程序执行

一个完整的程序是通过将一个名为mainpackage 的单个无转换包与它所导入的所有包传递链接而创建的.该main包中必须包名称main和声明函数main,它没有参数和返回值.

func main() { … }

程序执行从初始化main包然后调用该函数开始main.函数main返回时,程序退出.它不等待其他(非main)goroutines完成.

您没有向我们提供有关您问题的简单,简洁,可编译和可执行的示例.这是您的代码的精简版本.

package main

import (
    "fmt"
    "strconv"
    "sync"
    "time"
)

// structs
type Reading struct {
    Id   string
    Name string
}

var waitGroup sync.WaitGroup

func main() {
    // Setup timer
    startTime := time.Now()

    // Setup readings
    readings := prepareReadings()
    fmt.Println("readings prepared: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))

    // Insert readings
    for i := 1; i <= 1000000; i++ {
        waitGroup.Add(1)
        go insertReadings(readings)

        // fmt.Print(".")

        if i%100000 == 0 {
            fmt.Println("100000 readings queued for insert: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))
        }
    }
    waitGroup.Wait()

    fmt.Println("all readings inserted: " + strconv.FormatFloat(time.Since(startTime).Seconds(), 'f', 2, 64))
}

func insertReadings(readings []Reading) {
    waitGroup.Done()
}

func prepareReadings() []Reading {
    var readings []Reading
    for i := 1; i <= 1; i++ {
        readings = append(readings, Reading{Name: "Thing"})
    }
    return readings
}

输出:

readings prepared: 0.00
100000 readings queued for insert: 0.49
100000 readings queued for insert: 1.12
100000 readings queued for insert: 1.62
100000 readings queued for insert: 2.54
100000 readings queued for insert: 3.05
100000 readings queued for insert: 3.56
100000 readings queued for insert: 4.06
100000 readings queued for insert: 5.57
100000 readings queued for insert: 7.15
100000 readings queued for insert: 8.78
all readings inserted: 34.76

现在,逐个构建程序,并查看它开始失败的位置.